Timeline for Reference request: groups of multiplicative type are closed under extensions
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
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| Oct 11, 2015 at 23:57 | comment | added | Dave Witte Morris | That is correct. If we let $T$ be the group of diagonal matrices in $\mathrm{SL}_2(F)$, then the semidirect product is the subgroup $T \cdot \left\langle \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\right\rangle$ of $\mathrm{GL}_2(F)$. Everything is algebraic. | |
| Oct 11, 2015 at 19:52 | comment | added | მამუკა ჯიბლაძე | If I understand the question correctly, both $\langle\sigma\rangle$ and the semidirect product must be realized as linear algebraic groups, the former - of multiplicative type, and the maps in the extension as morphisms of algebraic groups, no? | |
| Oct 11, 2015 at 18:41 | history | answered | Dave Witte Morris | CC BY-SA 3.0 |